A295366 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3), where a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 2, 3, 12, 23, 44, 77, 132, 221, 367, 604, 987, 1608, 2613, 4240, 6873, 11134, 18029, 29186, 47240, 76453, 123720, 200201, 323950, 524181, 848162, 1372375, 2220570, 3592979, 5813584, 9406599, 15220220, 24626857

Offset

0,2

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Links

Table of n, a(n) for n=0..32.

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

Example

a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, so that
b(2) = 7 (least "new number")
a(3) = a(2) + a(1) + b(2) + b (1) - b(0) = 12
Complement: (b(n)) = (4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, ...)

Mathematica

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; b[0] = 4; b[1] = 5; b[2]=6;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] - b[n - 3];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 32; u = Table[a[n], {n, 0, z}]   (* A295366 *)
v = Table[b[n], {n, 0, 10}]  (* complement *)

Crossrefs

Cf. A001622, A295357.

Sequence in context: A122604 A282234 A024780 * A323120 A323121 A323122

Adjacent sequences: A295363 A295364 A295365 * A295367 A295368 A295369

Keyword

nonn,easy

Author

Clark Kimberling, Nov 22 2017

Status

approved